Answer
$0$
Work Step by Step
Here, we have $\lim\limits_{x \to 0^{+}} f(x)=\lim\limits_{x \to 0^{+}} \dfrac{\ln x}{1/x^2}$
But $\lim\limits_{x \to 0^{+}} \dfrac{\ln x}{1/x^2}=\dfrac{-\infty}{\infty}$
This shows an indeterminate form of limit, thus we will apply L-Hospital's rule such as:
$\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{p'(x)}{q'(x)}$
and $\lim\limits_{x \to 0^{+}} \dfrac{\frac{1}{x}}{\frac{-2}{x^3}}=0 $
